Difference between revisions of "1984 AIME Problems/Problem 8"

Revision as of 21:27, 26 March 2007

Problem

The equation $\displaystyle z^6+z^3+1$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in thet complex plane. Determine the degree measure of $\theta$ .

Solution

If $r$ is a root of $z^6+z^3+1$ , then $0=(r^3-1)(r^6+r^3+1)=r^9-1$ . The polynomial $x^9-1$ has all of its roots with absolute value 1 and argument of the form $40m^\circ$ for integer $m$ .

This reduces $\theta$ to either 120 or 160. But $\theta$ can't be 120 because if $r=\cos 120^\circ +i\sin 120^\circ$ , then $r^3=1$ and $r^6+r^3+1=3$ , a contradiction. This leaves $\theta=160$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-{{solution}}
+If <math>r</math> is a root of <math>z^6+z^3+1</math>, then <math>0=(r^3-1)(r^6+r^3+1)=r^9-1</math>. The polynomial <math>x^9-1</math> has all of its roots with absolute value 1 and argument of the form <math>40m^\circ</math> for integer <math>m</math>.
+This reduces <math>\theta</math> to either 120 or 160. But <math>\theta</math> can't be 120 because if <math>r=\cos 120^\circ +i\sin 120^\circ </math>, then <math>r^3=1</math> and <math>r^6+r^3+1=3</math>, a contradiction. This leaves <math>\theta=160</math>.
 == See also ==
 * [[1984 AIME Problems/Problem 7 | Previous problem]]
 * [[1984 AIME Problems/Problem 9 | Next problem]]
 * [[1984 AIME Problems]]

Page

Toolbox

Search

Difference between revisions of "1984 AIME Problems/Problem 8"

Revision as of 21:27, 26 March 2007

Problem

Solution

See also